Let X represent the time it takes from when someone enters the line for a roller coaster until they exit on the other side. Consider the probability model defined by the cumulative distribution function given below.
0 x < 3
F(x) = (x-3)/1.13 3 < x < 4.13
1 x > 4.13

What is E(X)? Give your answer to three decimal places.

What is the value c such that P(X < c) = 0.75? Give your answer to four decimal places.

What is the probability that X falls within 0.28 minutes of its mean? Give your answer to four decimal places.

a. E(x) = 3.730

b. c = 3.8475

c. 0.4308

Step-by-step explanation:

a.

Given

0 x < 3

F(x) = (x-3)/1.13, 3 < x < 4.13

1 x > 4.13

Calculating E(x)

First, we'll calculate the pdf, f(x).

f(x) is the derivative of F(x)

So, if F(x) = (x-3)/1.13

f(x) = F'(x) = 1/1.13, 3 < x < 4.13

E(x) is the integral of xf(x)

xf(x) = x * 1/1.3 = x/1.3

Integrating x/1.3

E(x) = x²/(2*1.13)

E(x) = x²/2.26 , 3 < x < 4.13

E(x) = (4.13²-3²)/2.16

E(x) = 3.730046296296296

E(x) = 3.730 (approximated)

b.

What is the value c such that P(X < c) = 0.75

First, we'll solve F(c)

F(c) = P(x<c)

F(c) = (c-3)/1.13= 0.75

c - 3 = 1.13 * 0.75

c - 3 = 0.8475

c = 3 + 0.8475

c = 3.8475

c.

What is the probability that X falls within 0.28 minutes of its mean?

Here we'll solve for

P(3.73 - 0.28 < X < 3.73 + 0.28)

= F(3.73 + 0.28) - F(3.73 + 0.28)

= 2*0.28/1.3 = 0.430769

= 0.4308 -- Approximated

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐